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# ch05

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```372
ro, 0, to 0.6357 0.3273 362.4 365
Continued \u2026..
PROBLEM 5.92 (Cont.)
The temperatures from the one-term series calculations of the Example 5.7 are systematically higher
than those resulting from the IHT multiple-term series model, which is the more accurate method.
(b) The temperature histories for the center and mid-height of the side locations are shown in the graph
below. Note that at early times, the temperature difference between these locations, and hence the
gradient, is large. Large differences could cause variations in microstructure and hence, mechanical
properties, as well as induce residual thermal stresses.
(c) Effect of doubling the convection coefficient is to increase the quenching rate, but much less than
by a factor of two as can be seen in the graph below.
Quenching with h = 500 W/m^2.K
0 2 4 6 8 10
Time, t (min)
300
400
500
600
T(x
, r,
t)
(C
)
Mid-height of side (0,ro)
Center (0, 0)
Effect of increased conv. coeff. on quenching rate
0 2 4 6 8 10
Time, t (min)
300
400
500
600
T(
0,
0,
t) (
C)
h = 500 W/m^2.K
h = 1000 W/m^2.K
COMMENTS: From IHT menu for Transient Conduction | Plane Wall and Cylinder, the models
were combined to solve the product solution. Key portions of the code, less the input variables, are
copied below.
// Plane wall temperature distribution
// The temperature distribution is
T_xtP = T_xt_trans(&quot;Plane Wall&quot;,xstar,FoP,BiP,Ti,Tinf) // Eq 5.39
// The dimensionless parameters are
xstar = x / L
BiP = h * L / k // Eq 5.9
FoP= alpha * t / L^2 // Eq 5.33
alpha = k/ (rho * cp)
// Dimensionless representation, P(x,t)
P_xt = (T_xtP - Tinf ) / (Ti - Tinf)
// Cylinder temperature distribution
// The temperature distribution T(r,t) is
T_rtC = T_xt_trans(&quot;Cylinder&quot;,rstar,FoC,BiC,Ti,Tinf) // Eq 5.47
// The dimensionless parameters are
rstar = r / ro
BiC = h * ro / k
FoC= alpha * t / ro^2
// Dimensionless representation, C(r,t)
C_rt= (T_rtC - Tinf ) / (Ti - Tinf)
// Product solution temperature distribution
(T_xrt - Tinf) / (Ti - Tinf) = P_xt * C_rt
PROBLEM 5.93
KNOWN: Stability criterion for the explicit method requires that the coefficient of the pmT
term of the one-dimensional, finite-difference equation be zero or positive.
FIND: For Fo > 1/2, the finite-difference equation will predict values of p+1mT which violate
the Second law of thermodynamics. Consider the prescribed numerical values.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in x, (2) Constant properties, (3) No
internal heat generation.
ANALYSIS: The explicit form of the finite-difference equation, Eq. 5.73, for an interior
node is
( ) ( )p+1 p p pm mm+1 m-1T Fo T T 1 2 Fo T .= + + \u2212
The stability criterion requires that the coefficient of pmT be zero or greater. That is,
( ) 11 2 Fo 0 or Fo .
2
\u2212 \u2265 \u2264
For the prescribed temperatures, consider situations for which Fo = 1, ½ and ¼ and calculate
p+1
mT .
( ) ( )
( ) ( )
( ) ( )
p+1
m
p+1
m
p+1
m
Fo 1 T 1 100 100 C 1 2 1 50 C 250 C
Fo 1/2 T 1 2 100 100 C 1 2 1/ 2 50 C 100 C
Fo 1/4 T 1/ 4 100 100 C 1 2 1/ 4 50 C 75 C.
= = + + \u2212 × =
= = + + \u2212 × =
= = + + \u2212 × =
/
\$
\$ \$
\$
\$ \$
\$
\$ \$
Plotting these distributions above, note that when Fo = 1, p+1mT is greater than 100°C, while
for Fo = ½ and ¼ , p+1mT \u2264 100°C. The distribution for Fo = 1 is thermodynamically
impossible: heat is flowing into the node during the time period \u2206t, causing its temperature to
rise; yet heat is flowing in the direction of increasing temperature. This is a violation of the
Second law. When Fo = ½ or ¼, the node temperature increases during \u2206t, but the
temperature gradients for heat flow are proper. This will be the case when Fo \u2264 ½, verifying
the stability criterion.
PROBLEM 5.94
KNOWN: Thin rod of diameter D, initially in equilibrium with its surroundings, Tsur,
suddenly passes a current I; rod is in vacuum enclosure and has prescribed electrical
resistivity, \u3c1e, and other thermophysical properties.
FIND: Transient, finite-difference equation for node m.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, transient conduction in rod, (2) Surroundings are
much larger than rod, (3) Properties are constant and evaluated at an average temperature, (4)
No convection within vacuum enclosure.
ANALYSIS: The finite-difference equation is derived from
the energy conservation requirement on the control volume,
Ac\u2206x, where 2cA D / 4 and P D.\u3c0 \u3c0= =
The energy balance has the form
p+1 p
2 m m
in out g st a b rad e
T TE E E E q q q I R cV .
t
\u3c1 \u2212\u2212 + = + \u2212 + =
\u2206
\ufffd \ufffd \ufffd \ufffd
where 2g eE I R=\ufffd and e e cR x/A .\u3c1= \u2206 Using Fourier\u2019s law to express the conduction terms,
qa and qb, and Eq. 1.7 for the radiation exchange term, qrad, find
( )p p p p p+1 pm m 4,p 4 2 e m mm-1 m+1c c m sur c
c
T T T T x T TkA kA P x T T I cA x .
x x A t
\u3c1
\u3b5 \u3c3 \u3c1
\u2212 \u2212 \u2206 \u2212
+ \u2212 \u2206 \u2212 + = \u2206
\u2206 \u2206 \u2206
Divide each term by \u3c1cAc \u2206x/\u2206t, solve for p+1mT and regroup to obtain
( )p+1 p p pm mm-1 m+12 2k t k tT T T 2 1 T
c cx x\u3c1 \u3c1
\uf8ee \uf8f9\u2206 \u2206
= \u22c5 + \u2212 \u22c5 \u22c5 \u2212\uf8ef \uf8fa
\u2206 \u2206\uf8f0 \uf8fb
( ) 24,p 4 em sur 2
c c
I P t tT T .
A c cA
\u3c1\u3b5 \u3c3
\u3c1 \u3c1
\u2206 \u2206
\u2212 \u22c5 \u2212 + \u22c5
Recognizing that Fo = \u3b1 \u2206t/\u2206x2, regroup to obtain
( ) ( ) ( ) 2 22p+1 p p p 4,p 4 em m m surm-1 m+1 2
c c
I xP xT Fo T T 1 2 Fo T Fo T T Fo.
kA kA
\u3c1\u3b5 \u3c3 \u2206\u2206
= + + \u2212 \u2212 \u22c5 \u2212 + \u22c5
The stability criterion is based upon the coefficient of the pmT term written as
Fo \u2264 ½. <
COMMENTS: Note that we have used the forward-difference representation for the time derivative;
see Section 5.9.1. This permits convenient treatment of the non-linear radiation exchange term.
PROBLEM 5.95
KNOWN: One-dimensional wall suddenly subjected to uniform volumetric heating and
convective surface conditions.
FIND: Finite-difference equation for node at the surface, x = -L.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional transient conduction, (2) Constant properties, (3)
Uniform \ufffdq.
ANALYSIS: There are two types of finite-difference equations for the explicit and implicit
methods of solution. Using the energy balance approach, both types will be derived.
Explicit Method. Perform an energy balance on the surface node shown above,
p+1 p
o o
in out g st conv cond
T TE E E E q q qV cV
t
\u3c1 \u2212\u2212 + = + + =
\u2206
\ufffd \ufffd \ufffd \ufffd \ufffd (1)
( )( ) ( ) p p p+1 pop o o1o T T T Tx xh 1 1 T T k 1 1 q 1 1 c 1 1 .
x 2 2 t
\u3c1
\u221e
\u2212
\u2212\u2206 \u2206\uf8ee \uf8f9 \uf8ee \uf8f9
\u22c5 \u2212 + \u22c5 + \u22c5 \u22c5 = \u22c5 \u22c5\uf8ef \uf8fa \uf8ef \uf8fa\u2206 \u2206\uf8f0 \uf8fb \uf8f0 \uf8fb\ufffd (2)
For the explicit method, the temperatures on the LHS are evaluated at the previous time (p).
The RHS provides a forward-difference approximation to the time derivative. Divide Eq. (2)
by \u3c1c\u2206x/2\u2206t and solve for p+1oT .
( ) ( )p+1 p p p po o o o12h t k t tT 2 T T 2 T T q T .
c x c
c x\u3c1 \u3c1\u3c1
\u221e
\u2206 \u2206 \u2206
= \u2212 + \u2212 + +
\u2206 \u2206
\ufffd
Introducing the Fourier and Biot numbers,
( ) 2Fo k/ c t/ x Bi h x/k\u3c1\u2261 \u2206 \u2206 \u2261 \u2206
( )
2p+1 p p
o o1
q xT 2 Fo T Bi T 1 2 Fo 2 Fo Bi T .
2k\u221e
\uf8ee \uf8f9\u2206
= + \u22c5 + + \u2212 \u2212 \u22c5\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb
\ufffd (3)
The stability criterion requires that the coefficient of poT be positive. That is,
( ) ( )1 2 Fo 2 Fo Bi 0 or Fo 1/2 1 Bi .\u2212 \u2212 \u22c5 \u2265 \u2264 + (4) <
Implicit Method. Begin as above with an energy balance. In Eq. (2), however, the
temperatures on the LHS are evaluated at the new (p+1) time. The RHS provides a backward-
difference approximation to the time derivative.
( ) p+1 p+1 p+1 pop+1 o o1o T T T Tx xh T T k q c
x 2 2 t
\u3c1
\u221e
\u2212
\u2212\u2206 \u2206\uf8ee \uf8f9 \uf8ee \uf8f9
\u2212 + + =\uf8ef \uf8fa \uf8ef \uf8fa\u2206 \u2206\uf8f0 \uf8fb \uf8f0 \uf8fb\ufffd (5)
( )( ) 2p+1 p+1 po o1 q x1 2 Fo Bi 1 T 2 Fo T T 2Bi Fo T Fo .k\u221e
\u2206
+ + \u2212 \u22c5 = + \u22c5 \u22c5 +
\ufffd
(6) <
COMMENTS: Compare these results (Eqs. 3, 4 and 6) with the appropriate expression in
Table 5.2.
PROBLEM 5.96
KNOWN: Plane wall,```